The header says it all. I "understand" the definition of a filtered category, in that I get what it's saying:

A category $J$ is filtered if

  1. $J$ is nonempty
  2. For a pair of objects $j, j' \in J$, there exists a third $k$ equipped with morphisms $u: j \to k$ and $v: j' \to k$.
  3. For a pair of parallel morphisms $u, v: i \to j$, there exists an object $k$ equipped with a morphism $w: j \to k$ such that $w \circ u = w \circ v$.

However, I do not truly understand because I don't see why anyone would ever come up with the specific definition. For example; why not make $w$ "unique" in (3)? Or perhaps remove the assumption of commutativity in (3)?

I've primarily studied from Mac Lane's text, but he doesn't give any motivation and there is also a lack of exercises when he discusses filtered categories. So I've been looking around a different texts, but they seem to all say the same thing: in introducing filtered categories, they say something dry along the lines of "filtered categories are generalizations of directed sets" which is then followed by the definition of a filtered category.

In reading more, it seems that these texts are trying to say that they generalize projective and inductive limits. However, I'm not seeing why we would need to do that. Is there something wrong with these constructions? A limiting assumption? If I'm looking at the limit of a functor $F: J \to \mathcal{C}$ where $J$ is a directed preorder, what benefit does regarding $J$ as filtered offer?

If this is just one of those things where I can only understand if I just shut up and keep reading, then let me know. I just personally prefer some motivation. Any help is appreciated! Note: this question is similar but not exactly what I'm after.

  • 3
    $\begingroup$ Filtered colimits are important largely because they commute with finite limits. This is important in topos theory, for example, to show why certain functors defined in terms of colimits preserve finite limits, which is a step to proving Diaconescu's theorem. $\endgroup$ Commented May 28, 2020 at 20:29
  • $\begingroup$ The point is that direct and inverse limits (what you're calling projective and inductive(? injective?) limits not necessarily in any order) are constrained by the assumption that the diagram shape is a directed preorder. Filtered categories generalize the directed assumption from preorders to all categories in such a way that as Malice Vidrine is saying above, the nice properties of direct limits generalize to all filtered colimits. (Similarly for inverse limits and cofiltered limits). $\endgroup$
    – jgon
    Commented May 28, 2020 at 23:55

1 Answer 1


I'm going to add a bit to my comment on the question, though this should be a self contained answer.


The idea is that direct limits (colimits over a directed preorder) are good and have nice properties that general colimits do not have, like commuting with finite limits. They are also much more computable and understandable compared to general colimits.

However, this restricts our source of diagrams to be directed preorders. But there are circumstances, when we wish to use more general domain categories, and we think that they should have similar properties.

For example, coverings of topological spaces form a directed preorder under refinement, if we just declare a covering $\mathcal{U}$ to refine $\mathcal{V}$ if for all $U\in\newcommand\U{\mathcal{U}}\U$ there is some $V\in\newcommand\V{\mathcal{V}}\V$ with $U\subseteq V$.

However, this is maybe not the best way to think about the category of open covers, depending on the situation. Instead, we might want to keep track of a particular choice of $V$ and inclusion map $U\subseteq V$ for each $U$. Now we have a category of coverings under refinement, which might fail to be a preorder. For example, if $\V=\{A,B\}$, and some $U\in\U$ is a subset of $A\cap B$, then there are at least two refinement morphisms from $\U\to \V$ (assuming there are any at all). However, we expect colimits over the refinement category to have the same nice property as colimits over our refinement preorder that we started with.

Thus we need to generalize the notion of directedness from preorders to all categories in such a way that it specializes to directedness for preorders, and ideally preserves the nice properties that we want.

Filtered Categories and Directed Preorders

This gives rise to the notion of filtered categories.

Recall that a directed preorder is a (nonempty) preordered set with the property that for any $x$ and $y$, there exists $z$ with $z\ge x$ and $z\ge y$.

These assumptions translate into requirements 1 and 2 of directed categories.

We make the (harmless) assumption that a directed category $J$ be nonempty. (It's harmless because we're only excluding one category, whose colimit we know is the initial object, so it hardly hurts to exclude this case, and it might make stating theorems easier).

Requirement 2 says that for any objects $j$ and $j'$ we can find an object $k$ with $u:j\to k$ and $v:j'\to k$. For a preorder this precisely reduces to for all $x$ and $y$ we can find $z$ with $x\le z$ and $y\le z$, since a morphism in a preorder from $x$ to $z$ exists exactly when $x\le z$, and similarly for $y$ and $z$.

Requirement 3 is the new requirement, but we'll notice that it is trivially satisfied by preorders, since there are never two distinct parallel morphisms.

Thus a preorder is filtered if and only if it is directed.

Understanding Requirement 3

Why then do we include requirement 3? Well, it says that $u$ and $v$ can be coequalized by some arrow. What does this give us? Well suppose we have a colimit of a diagram $X$ over a filtered category $J$ in $\mathbf{Set}$.

For each $j\in J$, we have a set $X_j$, and for each $u:j\to k$ in $J$, we have a function $u_*:X_j\to X_k$. We want to understand the colimit of $X$. We know that the colimit is the quotient of the disjoint union $\coprod_{j\in J} X_j$ under the equivalence relation generated by $x\sim u_*x$ for all $j,k\in J$, $u:j\to k$, and $x\in X_j$.

For directed limits, we know that we can identify this relation with the following: $x\sim y$ for $x\in X_j$, $y\in X_k$ if there is some $l\in J$ with $u:j\to l$ and $v:k\to l$ such that $u_*x=v_*y$. We'd like this to also be the case for general filtered categories.

Certainly this relation is always contained in the relation generated by $x\sim u_*x$, so we just have to prove that if $x\sim y$ in the colimit for $x\in X_j$, $y\in X_k$, then we can find such an $l$ and morphisms $u$ and $v$.

Suppose then that we have $x\sim y$. This means that we have a zig-zag of morphisms $$j=j_0\to j'_0 \leftarrow j_1\to j'_1 \leftarrow \cdots \to j'_{n-1} \leftarrow j_n=k$$ and elements $x_0,\ldots,x_n\in X_{j_0},\ldots,X_{j_n}$ such that pushing $x_i$ and $x_{i+1}$ to $X_{j'_i}$ gives the same result.

We want to show that in fact we can always take $n=1$, and we'll prove this by using our assumptions to reduce $n$ by $1$ when $n\ge 2$.

Take $j'_{n-2}$ and $j'_{n-1}$ and find some $j''$ with morphisms $j'_{n-2}\to j''$ and $j'_{n-1}\to j''$. We'd like to replace the $$j_{n-2}\to j'_{n-2}\leftarrow j_{n-1} \to j'_{n-1}\leftarrow j_n$$ part of our zig-zag with $$j_{n-2}\to j'_{n-2}\to j'' \leftarrow j'_{n-1}\leftarrow j_n,$$ which would give us a one shorter zig-zag, but we have a problem. We know pushing $x_{n-2}$ and $x_{n-1}$ to $j'_{n-2}$ gives the same result, and pushing $x_{n-1}$ and $x_n$ to $j'_{n-1}$ gives the same result, but what about pushing $x_{n-2}$ and $x_n$ to $j''$?

Well, we don't know. For $x_{n-2}$, this is the same as pushing $x_{n-1}$ to $j'_{n-2}$ and then to $j''$, and for $x_n$, this is the same as pushing $x_{n-1}$ to $j'_{n-1}$ and then to $j''$, but we don't know that these have the same result.

However, these are parallel maps from $j_{n-1}$ to $j''$, which means we can find some map from $j''$ to some $j^{(3)}$ which makes these two maps equal. Then if we use $j^{(3)}$ instead of $j''$, we do get a zig-zag that has length $n-1$, as desired.

This completes the proof, although admittedly, it may be very unclear, since I can't draw the pictures that I have in my head on this platform.

Final comment, a reinterpretation of the axioms

An equivalent set of requirements for a directed category $J$ is the following

  1. $J$ is nonempty
  2. For any finite diagram $X$ in $J$, there is a cocone.

This is because Requirement 2 in your version is essentially saying finite product diagrams have cocones and Requirement 3 is saying coequalizer diagrams have cocones. Putting these together, the same proof idea as (binary coproducts + coequalizers = finitely cocomplete) gives that all finite diagrams have cocones.

This also drastically simplifies the final part of my proof above. We can just take a cocone to the zig-zag, and it will automatically be the objects and morphisms we are looking for.

  • 1
    $\begingroup$ This is exactly what I was looking for! This makes sense, but I will have to draw out the diagrams myself to fully understand. However this definitely makes me more comfortable with filtered limits; thank you so much! $\endgroup$
    – trujello
    Commented May 30, 2020 at 0:07

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